B. Hongcow Solves A Puzzle

http://codeforces.com/contest/745/problem/B

题目要求的是，给定一个图形，要求里面判断是否有矩形，且仅有一个

就是

XXX....

XXX...X

是不行的，因为有两个了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
int n, m;
const int maxn =  + ;
char str[maxn][maxn];
char sub[maxn][maxn];
int tonext[][] = {{, }, {, }, {, -}, {-, }};
bool vis[maxn][maxn];
void dfs(int x, int y) {
    for (int i = ; i < ; ++i) {
        int tx = x + tonext[i][];
        int ty = y + tonext[i][];
        if (tx >=  && tx <= n && ty >=  && ty <= m && !vis[tx][ty] && str[tx][ty] == 'X') {
            vis[tx][ty] = true;
            dfs(tx, ty);
        }
    }
}
bool check() {
    for (int i = ; i <= n; ++i) {
        for (int j = ; j <= m; ++j) {
            if (str[i][j] == 'X' && !vis[i][j]) return false;
        }
    }
    return true;
}
void work() {
    scanf("%d%d", &n, &m);
    for (int i = ; i <= n; ++i) {
        scanf("%s", str[i] + );
    }
    int ans = ;
    for (int i = ; i <= n; ++i) {
        bool flag1 = false;
        bool flag2 = false;
        bool be = false;
        for (int j = ; j <= m; ++j) {
            if (str[i][j] == 'X') {
                if (str[i - ][j] == 'X') {
                    if (be && !flag1) {
                        cout << "NO" << endl;
                        return;
                    }
                    flag1 = true;
                }
                if (str[i + ][j] == 'X') {
                    if (be && !flag2) {
                        cout << "NO" << endl;
                        return;
                    }
                    flag2 = true;
                }
                be = true;
            }
            if (str[i][j] == 'X' && flag1) {
                if (str[i - ][j] != 'X') {
                    cout << "NO" << endl;
                    return;
                }
            }
            if (str[i][j] == 'X' && flag2) {
                if (str[i + ][j] != 'X') {
                    cout << "NO" << endl;
                    return;
                }
            }
        }
    }
//    cout << "ff" << endl;
    for (int i = ; i <= n; ++i) {
        for (int j = ; j <= m; ++j) {
            if (str[i][j] == 'X') {
                vis[i][j] = true;
                dfs(i, j);
                if (check()) {
                    cout << "YES" << endl;
                } else cout << "NO" << endl;
                return;
            }
        }
    }
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return ;
}

感觉我判断起来有点麻烦啊